$y=mx$ and $ax+by+c=0$ are perpendicular and meet at $(-9,6)$. Find the area of a related triangle.

Question

I was trying to solve this problem, but I'm not sure how to approach this problem. Here is the problem:

The lines $y = mx$ and $ax + by +c = 0$ are perpendicular to each other, and they intersect at $P(-9, 6)$. Find the area of the triangle $APO$. (Figure is not drawn to scale.)

Answer 1

Hints:

You just need to find the $x-$coordinate of point $A$. To get that you need the equation of the line $PA$. To get line $PA$ you need two pieces of information: namely a point (in your case $P$ is given) and the slope. Line $OP$ and line $PA$ are perpendicular. Slope of $OP$ is $\frac{6-0}{-9-0}=-\frac{2}{3}$. So slope of $PA$ is $\frac{3}{2}$. Now find the equation and once you have the point $A$, area can be computed easily ($\frac{1}{2}bh$).

Answer 2

$P = (-9,6)\\ A = (-9,6) + \frac 23(-6,-9) = (-13,0)$

$39$

or

One leg has length $\sqrt {9^2+6^2}$ the other is $\frac 69\sqrt {9^2+6^2}$

$\frac 12 \frac 69 (9^2 + 6^2) = 39$